Simplify the following expression and state the condition under which the simplification is valid. You can assume that $n \neq 0$. $z = \dfrac{16n - 56}{6} \times \dfrac{-7}{n(2n - 7)} $
When multiplying fractions, we multiply the numerators and the denominators. $z = \dfrac{ (16n - 56) \times -7 } { 6 \times n(2n - 7) } $ $ z = \dfrac {-7 \times 8(2n - 7)} {6 \times n(2n - 7)} $ $ z = \dfrac{-56(2n - 7)}{6n(2n - 7)} $ We can cancel the $2n - 7$ so long as $2n - 7 \neq 0$ Therefore $n \neq \dfrac{7}{2}$ $z = \dfrac{-56 \cancel{(2n - 7})}{6n \cancel{(2n - 7)}} = -\dfrac{56}{6n} = -\dfrac{28}{3n} $